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I am trying to understand an example of semidirect products that I read in Dummit and Foote. This is the example of the classification of groups of order 12 in page 182 and 183. In this case we consider that $G$ is a group of order 12, $V \in Syl_2(G)$ and $T\in Syl_3(G)$. My first question comes when we are assuming $V\trianglelefteq G$and $V \cong Z_2 \times Z_2$. I understand why we have 1 trivial homomorphism from $T \rightarrow Aut(V)$ and 2 non-trivial.

Q1) Why, when we consider these two non-trivial homomorphisms, do we get the group $A_4$? I know that the books makes a reference to a previous example, but I still don't understand where we get this group $A_4$ from. Can anyone explain this?

Also, I have a question when we assume $T \trianglelefteq G$. Particularly, I have a question when we are assuming $V \cong Z_2 \times Z_2$. I can see why we have three nontrivial homomorphisms.

Q2) What I am failing to see is why the resulting three semidirect products are all isomorphic to $S_3 \times Z_2$, can anyone explain this?

Q3) In general, I am having trouble, once I identify the possible homomorphisms into the automorphism of some subgroup of a group, identifying what groups the resulting semidirect product could be. Can anyone suggest some way to see this, or give any hint on how to do it? Thanks so much for your help!

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Let address your questions one by one.

1) Why, when we consider these two non-trivial homomorphisms, do we get the group $A_4$?

In the book the authors show that you have only 3 possible homomorphisms for the given semidirect product, two of which produce the same group.

So you have only two possible semidirect products one of which is the direct product.

Since $A_4$ is a semidirect product of type considered but it is not abelian (i.e. it is not isomorphic to the direct product) it must be the only other possible semidirect product, the only non abelian one.

2) What I am failing to see is why the resulting three semidirect products are all isomorphic to $S_3 \times Z_2$, can anyone explain this?

Using the notation of the book, let $a$ and $b$ denote the generators of $V$. If $\ker \varphi=\langle a \rangle$ it means that in the corresponding semidirect product $\langle a\rangle$ is normal while $\langle b\rangle$ is not.

This implies that $V \langle b \rangle$ is a subgroup of $G$ which is not abelian, it has order 6 and so it must be isomorphic to the unique non abelian group of order $6$, i.e. $S_3$.

This subgroup has index 2, hence it is normal and so applying the theorem for direct products to the normal subgroups $V\langle b \rangle\cong S_3$ and $\langle a \rangle \cong Z_2$ you get the wished isomorphism.

3) In general, I am having trouble, once I identify the possible homomorphisms into the automorphism of some subgroup of a group, identifying what groups the resulting semidirect product could be. Can anyone suggest some way to see this, or give any hint on how to do it?

I do not know of any general technique to do that. The point is that for doing what you say, at least if I have understood you correctly, one would need to know all the groups of a given cardinality in order to establish the isomorphism type of a direct product.

Unfortunately there is complete classification of all groups. There is a monster theorem which classifies all finite simple groups but we do not know all the possible group operations neccesary to generate all groups from these basic block.

I hope this address all your doubts.

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